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3 years ago

Alcohol begins to affect your brain when

1 answer:

7 0

<span>Alcohol begins to affect your brain when it crosses the blood-brain barrier. After this, it acts on the nerve cells and disrupts their communication with each other as well as other body parts. It inhibits the activity of some neural pathways due to which a person starts feeling lethargic, sluggish, and slow-moving.</span>

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A 12.0 V car battery is being used to power the headlights of a car. Each of the two headlights has a power rating of 37.7 Watts

marshall27 [118]

Answer:

3.921*10^19 electrons

Explanation:

To find the number of electron that trough the car battery you first calculate the current by using the following formula, which relates the power with the voltage:

$P=IV$ (1)

P: power of both headlights = 2(37.7W) = 75.4 W

I: current = ?

V: voltage of the battery = 12.0 V

You solve the equation (1) for I:

$I=\frac{P}{V}=\frac{75.4W}{12.0V}=6.283A=6.238\frac{C}{s}$

Next, you use the equivalence 1C = 6.241*10^18 electrons:

$6.238\frac{C}{s}*\frac{6.241*10^{18}\ electrons}{1C}=3.921*10^{19}\ electrons/s$

The number of electrons that cross the battery per second is 3.921*10^19 electrons

6 0

3 years ago

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

$\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}$

Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

$\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz$

8 0

3 years ago

Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec

scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0

4 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for t

egoroff_w [7]

Answer:

a. 5 batteries b. 1050 mAh

Explanation:

Here is the complete question

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?

Solution

A) How many batteries are needed?

Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries

So, the number of batteries needed is 5.

B) What mAh capacity should the batteries have?

Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.

So, Q = It = 150 mA × 7 h = 1050 mAh.

So, the batteries should have a mAh of 1050 mAh

4 0

3 years ago